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Let $A$ be a $2\times 2$ matrix with $tr(A) > 2$. Prove that for any positive integer $n$, $A^n ≠ I$.

I feel like I should approach this with respect to eigenvalues, i.e. the sum of the eigenvalues of $A$ is greater than $2$. However, I don't know where to head from here. Any help or guidance to a direction would be greatly appreciated!

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linearalgebrathrowaway
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3 Answers3

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First note that if $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$. As you pointed out, the trace being greater than $2$ implies that the sum of the eigenvalues is greater than $2$, so at least one eigenvalue $\lambda$ of $A$ satisfies $|\lambda|>1$. Thus $A^n$ has an eigenvalue $\lambda^n$, and $|\lambda^n|=|\lambda|^n>1$. This implies that $A^n\neq I$.

TomGrubb
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    Ah of course. Should be fixed if I replace $\lambda>1$ with $|\lambda|>1$, correct? – TomGrubb May 20 '15 at 03:36
  • Your usage of absolute value sign threw me off a little, why did you need to replace λ>1 with |λ|>1? Sorry for being nitpicky! – linearalgebrathrowaway May 20 '15 at 03:52
  • No problem! At first I forgot that the matrix need not have any real eigenvalues. Thus if $\lambda$ is complex, the expression $\lambda>1$ makes no sense. What we really care about is just that it has a strictly larger magnitude than $1$, so that $\lambda^n$ has a strictly larger magnitude than $1$. – TomGrubb May 20 '15 at 03:54
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Recall the characteristic equation $$ A^2-tr(A)A+I_2\det A=O_2.$$ Assume by contradiction that, for some $n\ge1$, $A^n=I$. Then $\det A=\pm1$, $A$ is invertible with inverse $A^{n-1}$, and $n\neq1,2$ because $tr(A)>2$. So $n\ge3$.

Multiply the characteristic equation by $A^{n-2}$ to get $ I-tr(A)A^{n-1}+\det A\,A^{n-2}=O_2$, $A^{n-1}(tr(A)I-\det A\,A)=I$, that is, $tr(A)I-\det A\,A=A$ from which you can get several contradictions.

Neutral Element
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First note that if λλ is an eigenvalue of AA, then λnλn is an eigenvalue of AnAn. As you pointed out, the trace being greater than 22 implies that the sum of the eigenvalues is greater than 22, so at least one eigenvalue λλ of AA satisfies |λ|>1|λ|>1. Thus AnAn has an eigenvalue λnλn, and |λn|=|λ|n>1|λn|=|λ|n>1. This implies that An≠IAn≠I.

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